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A drum of 40 cm radius has a capacity of 440 dm^(3) of water. Itcontains 396 dm^(3) of water and is placed on a solid block of exactly the same size as of drum. If a small hole is made at lower end of the drum perpendicular to its length, find the horizontal range of water on the ground in the beginning. Given g=10m//s^(2) |
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Answer» Solution :Radius of drum, `r=40cm` VOLUME of the drum `=440dm^(3)` `=4.4xx10^(5)cm^(3)` `[thereforedm=0.1m=10cm]` If h is the HEIGHT of the drum, then `pir^(2)h=4.4xx10^(5)` (or)`h=(4.4xx10^5)/((22//7)xx40xx40)` `=87.5cm`Height of block, `h_(1)=h=87.5cm` Volume of water `=396dm^(3)` `=3.96xx10^(5)cm^(3)` Let `h_(2)` be the height of the water column in the drum. Then `pir^(2)h^(2)=3.96xx10^(5)` `h_(2)=(3.96xx10^(5))/((22//7)xx(40)^(2))=78.75cm` Time taken by water to reach the GROUND, `t=sqrt((2h_1)/(g))` Velocity of efflux of water, `V=sqrt(2gh_2)` Horizontal range, `R=Vt=sqrt(2gh_2)xxsqrt((2h_1)/(g))` `=2sqrt(h_(1)h_(2))` `=2sqrt(87.5xx78.75)=166cm` |
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